Talk:Baker's map

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Attractor[edit]

The baker's map, as defined, has the entire square as an attractor, so the Hausdorff dimension is 2. The action of the baker's map is to squeeze the left half into the lower half and the right half into the upper half after a flip. The horseshoe map has an attractor that is the product of two Cantor sets. The map as defined in the horseshoe page has an attractor with fractal dimension –log(4) / log(a'). The minus sign is because a is smaller than 1 (as it needs to be smaller than 1/2). So if a = 0.45, the fractal dimension of the attractor of the horseshoe would be approximately 1.736.   XaosBits 21:58, 1 Jun 2005 (UTC)

...Attractor[edit]

Oh yeah, thanks, I was thinking that the map of which we were speaking was the bakers map with = 1/4, which would be 1 1/2. Thanks. In fact, the way I learned it the constant y is multiplied by has to be less than 1/2, so I was like Hey, that can't be 2...

formula[edit]

When the formula changed to "human readable format" I think it also changed into the horseshoe map, because of the -y.--MarSch 09:44, 23 Jun 2005 (UTC)


To flip or not to flip[edit]

The definition of the baker's map needs to include a flip of the right column to match the horseshoe map. Some authors do not include the flip and others do not:

  • Halmos: doesn't flip
  • Ott: doesn't flip
  • Cvitanovic: flips
  • Farmer: doesn't flip
  • Balazs & Voros: flips
  • Arnold & Avez: flips (2nd hand)

If you are working just with the baker's map, it's simpler not to flip. But if you are using the baker's map as some simplification of a smooth map, then you usually need to flip. Halmos and Arnold are the older references and as far as I can tell the origin of the citation chain (tree?). XaosBits 13:44, 23 Jun 2005 (UTC)

Yes, but the eigenfunctions of the tent map are ickier to understand; the symmetry about x=1/2 causes a degeneracy. The eigenfunctions of the non-flipped Bernoulli map are the Bernoulli polynomials. Also, if you are interested in eigenfunctions of Baker's map on the torus/elliptic curve, the non-flipped version makes more sense. (I think the flipped version gives the Klein bottle or something like that; I'm trying to understand this now). At any rate, the two maps are not equivalent, they behave differently; maybe this is what should be said. linas 00:04, 24 Jun 2005 (UTC)
Agreed. The eigenvalues for the tent map are just the even powers of 1/2 (for analytic functions), but I have no idea about the eigenfunctions. I have no strong opinions about reverting the text the way Linas had it or leaving it the way it is. I changed the article and only after that did I decided to poll the literature. If I had polled first, I would have left it as it was. XaosBits 04:14, 24 Jun 2005 (UTC)
The Bernoulli map was changed to the tent map, but it's description was not changed. It is clear that the Bernoulli map lops off dyadic digits, so the tent map does something else. What is the section below supposed to indicate? --MarSch 12:49, 24 Jun 2005 (UTC)

There is an article tent map but its a stub. The eigenvalues of the tent map of unit height are linear combos of Bernoulli. The interested reader is directed to the reference. I'd expand these articles, except for the fact that I am doing original research in very close proximity to this topic area, so whatever I wrote would be strongly colored. I'm happy to let it be. linas 18:16, 24 Jun 2005 (UTC)

that doesn't address my observations at all.--MarSch 18:47, 24 Jun 2005 (UTC)
Is there a problem? Whats the problem? linas 22:56, 24 Jun 2005 (UTC)
The tent map also lops off dyadic digits from the symbolic representation of a trajectory. Given a hyperbolic map (the tent map is hyperbolic), it always possible to set up symbolic representation for trajectories of a map (and flows too). The symbolic representation is usually set up by considering the inverse of the map. (An example is given in the horseshoe map article.) XaosBits 23:35, 24 Jun 2005 (UTC)
My concern is that the tent map and the Bernouilli map are not the same and since the Bernouilli map lops off digits, the tent map does not. I think you are saying that you could define some alternative representation of the unit interval by diadic expansions to make the tent map lop off digits and no more. This doesn't seem a very useful way of looking at it to me. If you think it is you should explain it better.--MarSch 10:28, 25 Jun 2005 (UTC)
The representation is not of the interval, but of the orbits. See sec 11.7 and chapter 12 of Chaos Book or other references cited in the dynamical systems article. XaosBits 13:06, 25 Jun 2005 (UTC)
Orbits of unit-height tent map

Well, yes, there is indeed an interesting interplay. If one has a string of dyadic digits, it can be represented by a real number, as binary; this is true both for Bernoulli map orbits and tent map orbits. For the bernoulli map, one has a very direct correspondance between the value of x and the binary representation of the orbit -- indeed they are one and the same. For the tent map, they are not. Hmm. Come to think of it, I have never seen a graph of the binary values of the orbits of the tent map, as a function of x. I'll create a few graphs now, and post shortly. BTW, XaosBits, do you perchance have any contact with Andreas Wirzba (who I think is a coauthor of the chaos book?) I wanted to say hi to him, we went to school together. linas 16:20, 25 Jun 2005 (UTC)

OK, see image, click on it to get detailed explanation. linas 18:11, 25 Jun 2005 (UTC)

Bernoulli map[edit]

(This was the section without the flip)

The non-fliping baker's map acts on the phase space of the unit square as

The non-fliping baker's map is a two-dimensional analog of the Bernoulli map

The Bernoulli map can be understood as the map that progressively lops digits off the dyadic expansion of x. Unlike the Bernoulli map, the baker's map is invertible.

As with many deterministic dynamical systems, the baker's map is studied by its action on the space of functions defined on the unit square. The Baker's map defines an operator on the space of functions, known as the transfer operator of the map. The Baker's map is an exactly solvable model of deterministic chaos, in that the eigenfunctions and eigenvalues of the transfer operator can be explicitly determined.

The non-singular eigenfunctions of the Bernoulli map are the Bernoulli polynomials. The full eigenfunction spectrum is given by the Hurwitz zeta function. A set of fractal eigenfunctions can be given as well. The eigenfunctions of the baker's map are the Lévy C curve.

Baker's map image[edit]

The russian version of this article has a nice image of a baker, which should be copied to wiki commons and then included here. I tried to include the image directly from the russian site, but the inclusion did not work.

The russian article also has a reatment of the symbolic dynamics which should be copied here. linas 19:15, 24 September 2006 (UTC)[reply]

\mathrm is not the same as \text (nor is \mbox)[edit]

I found this:

I changed it to this:

The first uses \mathrm{baker-folded}; the second uses \text{baker-folded}. When \mathrm is used, the hyphen becomes a minus sign; when \text is used, it remains a hyphen. The first uses \mbox{for }; the second uses \text{for }. In some contexts, those look much more different from each other than in the example above. For example, contrast \min_\mbox{abcd} with \min_\text{abcd}:

The purpose of \mbox is to prevent line-breaks when TeX is used in the usual way as opposed to the way it's used within Wikipedia. It shouldn't be used as a substitute for \text. Michael Hardy (talk) 15:56, 18 February 2010 (UTC)[reply]

BakedBill.jpg[edit]

I removed this image because it didn't demonstrate the folded bakers map (or indeed the unfolded bakers map) as it was defined in the article. As evidence to this fact, note how the folded baker's map (as defined) would map a point arbitrarily close to (1, 1) to a point arbitrarily close to (0, 1/2) and the unfolded version would take such a point near to (1, 1), whereas this image mapped the top right corner (1, 1) to the top left (0, 1). Also note that both bakers maps preserve chirality (clock-wise gets mapped to clockwise) in regions they don't split whereas this map turned the right 5 into it's (squashed) mirror image. Can someone please upload a correct demonstration as it was a good illustration aside from the fact it was wrong. Nbrader (talk) 13:51, 18 April 2014 (UTC)[reply]